Problem: The lifespans of tigers in a particular zoo are normally distributed. The average tiger lives $26.4$ years; the standard deviation is $4$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a tiger living less than $38.4$ years.
Answer: $26.4$ $22.4$ $30.4$ $18.4$ $34.4$ $14.4$ $38.4$ $99.7\%$ $0.15\%$ $0.15\%$ We know the lifespans are normally distributed with an average lifespan of $26.4$ years. We know the standard deviation is $4$ years, so one standard deviation below the mean is $22.4$ years and one standard deviation above the mean is $30.4$ years. Two standard deviations below the mean is $18.4$ years and two standard deviations above the mean is $34.4$ years. Three standard deviations below the mean is $14.4$ years and three standard deviations above the mean is $38.4$ years. We are interested in the probability of a tiger living less than $38.4$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $99.7\%$ of the tigers will have lifespans within 3 standard deviations of the average lifespan. The remaining $0.3\%$ of the tigers will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({0.15\%})$ will live less than $14.4$ years and the other half $({0.15\%})$ will live longer than $38.4$ years. The probability of a particular tiger living less than $38.4$ years is ${99.7\%} + {0.15\%}$, or $99.85\%$.